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Math Help - Limits question

  1. #1
    Member akhayoon's Avatar
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    Limits question

    Limit question on a midterm that I have and still haven't the slightest idea on how to complete

    Find all Values of a and b such that \lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1

    all I could figure out was that if you use L'Hopitale's rule you would get

    \lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1

    but I have no idea how to use this to solve for any variables

    thank you
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  2. #2
    Moo
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    Hello,

    Why not using a second time l'H˘pital's rule ?
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  3. #3
    Member akhayoon's Avatar
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    because it would be (0/0) ?
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  4. #4
    Moo
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    Hm ok, I just don't know correctly the conditions for applying this rule
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Hi

    Another approach : \lim_{x\to 0} \frac{\sin(2x)-\sin(0)}{2x-0}=\cos 0 = 1
    Hence, \lim_{x \to 0} \frac{3}{x}\frac{\sin(2x)}{x}+a-\frac{b}{x}=\lim_{x \to 0} \frac{6}{x} \frac{\sin(2x)}{2x}+a-\frac{b}{x}=\lim_{x\to 0}\frac{6-b}{x}+a=1
    When is this possible ?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    Limit question on a midterm that I have and still haven't the slightest idea on how to complete

    Find all Values of a and b such that \lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1

    all I could figure out was that if you use L'Hopitale's rule you would get

    \lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1

    but I have no idea how to use this to solve for any variables

    thank you
    L'hopital it again if you wish...it works...but look...you have two equations with two variables...solve for them using simulataneous equations
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  7. #7
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    Quote Originally Posted by akhayoon View Post
    Limit question on a midterm that I have and still haven't the slightest idea on how to complete

    Find all Values of a and b such that \lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1

    all I could figure out was that if you use L'Hopitale's rule you would get

    \lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1

    but I have no idea how to use this to solve for any variables

    thank you
    Obviously you require another indeterminant form. This will happen if b = -6. So then you have:

    \lim_{x \rightarrow 0} \frac{6 \cos(2x) + 2ax - 6}{2x}=1.

    Apply l'Hopital again:

    \lim_{x \rightarrow 0}\frac{-12 \sin (2x) + 2a}{2}=1.

    In other words, \frac{2a}{2} = 1 \Rightarrow a = 1.
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  8. #8
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    Quote Originally Posted by akhayoon View Post
    Limit question on a midterm that I have and still haven't the slightest idea on how to complete

    Find all Values of a and b such that \lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1

    all I could figure out was that if you use L'Hopitale's rule you would get

    \lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1

    but I have no idea how to use this to solve for any variables

    thank you
    All your troubles would have been avoided of course if you hadn't run to the l'Hospital with a sore thumb. Try substituting the series expansion around x = 0 for sin(2x):


    \Rightarrow \lim_{x \rightarrow 0} \frac{3 (2x - [2x]^3/6 + ...) + ax^{2} - bx}{x^2}=1


    \Rightarrow \lim_{x \rightarrow 0} \frac{(6x - 4x^3/3 + ...) + ax^{2} - bx}{x^2}=1


    \Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) x + ax^{2} - 4x^3/3  + ...}{x^2}=1


    \Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) + ax - 4x^2/3  + ...}{x} = 1

    There's a problem unless 6 - b = 0 => b = 6. Then:


    \Rightarrow \lim_{x \rightarrow 0} \frac{ax - 4x^2/3  + ...}{x} = 1


    \Rightarrow \lim_{x \rightarrow 0} \left(a - \frac{4x}{3}  + ...\right) = 1

    Therefore a = 1.

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